Event Horizon – Meaning of Photon or Light Being ‘Frozen’ at the Event Horizon

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Objects on this site are said to "freeze" at the event horizon, but what about photons themselves?

I have read this question:

At the event horizon $v_{eff} = 0$ and the light beam is frozen at the horizon unable to move outwards.

If you shoot a light beam behind the event horizon of a black hole, what happens to the light?

And this one:

At large distances (large r) the velocity tends to 1 (i.e. c) but close to the black hole it decreases, and falls to zero at the event horizon.
But, but, but, be absolutely clear what you're calculating. All you're calculating is the speed of light in your co-ordinate system i.e. the result you get applies only to you. Other observers in other places will calculate a different value, and every observer everywhere will find the local speed of light to have the same value of c.

Speed of light in a gravitational field?

Now two very important rules about photons/light that come to mind:

  1. photons (or classically light) do not have a rest frame (special relativity)

  2. photons (or light) can never be brought to rest (they move at speed c always)

The speed of photons is always c, when measured locally in vacuum. This question is about a non-local measurement.

However, when I measure the speed of light based on the second answer above, I get zero at the event horizon, which might be interpreted, as being at rest, that is having a rest frame (from an external point of view).

Question:

  1. What does it exactly mean when we say that a photon or light is "frozen" at the event horizon?

Best Answer

It doesn't mean a lot.

Event horizons are outward-moving lightlike surfaces, but the curvature is such that usually their overall size doesn't increase (except during black hole formation, accretion of matter, and mergers).

If you cover a part of the event horizon with local inertial coordinates, the speed of the light on the horizon is $c$ with respect to those coordinates. But there isn't any inertial or approximately inertial coordinate system that covers the entire horizon (because of spacetime curvature). Some common coordinate systems that do cover the whole horizon use an $r$ coordinate that is constant on the horizon (when it isn't expanding). The $dr/dt$ speed of outward-moving light on the horizon is therefore zero. The same light at the same time has a speed of $c$ in the inertial coordinates. The light isn't at rest. It's moving at the speed of light, which just happens to be $0$ in certain coordinate systems.

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